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Consider the following relational schema: Doctor(DName,Reg_no) Patient(Pname, Disease) Assigned_To (Pname,Dname) Give expression in both Tuple calculus and Domain calculus for every of the queries:(i) Get the names of patients who are assigned to more than one doctor. (ii) Get the names of doctors who are treating patients with ‘Polio'.
Ans: (i)Tuple Calculus: {p[PName] | p ∈ PATIENT ∧ ∃a1, a2 (a1 ∈ ASSIGNED_TO ∧ a2 ∈ ASSIGNED_TO ∧p[PName] = a1[PName] ∧ a1[PName] = a2[PName] ∧ a1[DName] ≠ a2[DName])} Domain Calculus:{p | ∃p1, d1, p2, d2 (
∈ PATIENT ∧ ∈ ASSIGNED_TO ∧ ∈ASSIGNED_TO ∧ p1 = p2 ∧ d1 ≠ d2)} (ii)Tuple Calculus: {u[Dname] | u ∈ ASSIGNED_TO ∧ ∃t (t ∈ PATIENT ∧ t[Disease] = ‘Polio' ∧t[PName] = u[PName])} Domain Calculus: {d | ∃p1, p2, s2 ( ∈ ASSIGNED_TO ∧ ∈ PATIENT∧ p1 = p2 ∧ s2 = ‘Polio')}
What is 2NF? A relation schema R is in 2NF if it is in 1NF and every non-prime attribute A in R is fully functionally dependent on primary key.
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